<h2>Problem 132</h2>
<div style="color:#666;font-size:80%;">01 December 2006</div><br />
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<p>A number consisting entirely of ones is called a repunit. We shall define R(<i>k</i>) to be a repunit of length <i>k</i>.</p>
<p>For example, R(10) = 1111111111 = 11<img src='images/symbol_times.gif' width='9' height='9' alt='&times;' border='0' style='vertical-align:middle;' />41<img src='images/symbol_times.gif' width='9' height='9' alt='&times;' border='0' style='vertical-align:middle;' />271<img src='images/symbol_times.gif' width='9' height='9' alt='&times;' border='0' style='vertical-align:middle;' />9091, and the sum of these prime factors is 9414.</p>
<p>Find the sum of the first forty prime factors of R(10<img src="" style="display:none;" alt="^(" /><sup>9</sup><img src="" style="display:none;" alt=")" />).</p>

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